Question

Find the inverse of the function,$y = {3^x}$ and check if the inverse is also a function.

Answer 1

Hint:Inverse of a given one to one function $f(x)$ is defined as, $g(x)$. Here $g(x)$ is equal to ${f^{ - 1}}(x)$. We may also say that if $f(x)$ and $g(x)$ are two one-to-one functions, then $(fog)(x) = x$. Also, to be kept in mind that, \[{f^{ - 1}}(x) \ne \dfrac{1}{{f(x)}}\].

Complete step by step solution:
Let us assume, $f(x) = y$.
And, we know from the question that $y = {3^x}$
Replace every x with y and every y with x.
Thus, $f(y) = x = {3^y}$……. equation(i)
Solving for y, we have
\[
\log (f(y)) = \log {3^y} \\
\Rightarrow \log (x) = y\log 3 \\
\Rightarrow y = \dfrac{{\log (x)}}{{\log 3}} \\
\]
The resultant function in y be named ${f^{ - 1}}(x)$.
This function ${f^{ - 1}}(x)$ is the inverse of the function, $f(x)$. To verify the same, we need to put the values in these equations and tally the answers.
\[\left( {f \circ {f^{ - 1}}} \right)\left( x \right) = x\]…….. equation(iii)
\[\left( {{f^{ - 1}} \circ f} \right)\left( x \right) = x\]…….. equation(iv)
Equation (iii) can be written as,
\[\left( {f \circ {f^{ - 1}}} \right)\left( x \right) = f[{f^{ - 1}}(x)]\]
Putting the values of ${f^{ - 1}}(x)$ in the above equation,
\[ \Rightarrow \left( {f \circ {f^{ - 1}}} \right)\left( x \right) = f[\dfrac{{\log (x)}}{{\log 3}}]\]
\[ \Rightarrow \left( {f \circ {f^{ - 1}}} \right)\left( x \right) = {3^{\dfrac{{\log (x)}}{{\log 3}}}} =
x\]……….(Hence the equation is verified.)
Next, we need to verify equation (iv),
\[\left( {{f^{ - 1}} \circ f} \right)\left( x \right) = {f^{ - 1}}[f(x)]\]
Putting the values of in the above equation,
\[ \Rightarrow \left( {{f^{ - 1}} \circ f} \right)\left( x \right) = {f^{ - 1}}[{3^x}]\]
\[ \Rightarrow \left( {{f^{ - 1}} \circ f} \right)\left( x \right) = \dfrac{{\log [{3^x}]}}{{\log 3}} =
x\]……….. (Hence the equation is verified.)
As the definition of an inverse function says, two one-to-one functions are inverse of each other if they satisfy the equations \[\left( {f \circ {f^{ - 1}}} \right)\left( x \right) = x\]and \[\left( {{f^{ - 1}} \circ f} \right)\left( x \right) = x\], and hence the found inverse of the given function is also a function.

Note: Let $f$ be a function whose domain is called set X, and whose codomain is called set Y. Then $f$ is invertible if there exists a function g with domain Y and image (range) X, with the property: $f(x) = y{\mkern 1mu} {\mkern 1mu} \Leftrightarrow {\mkern 1mu} {\mkern 1mu} g(y) = x.$
If $f$ is invertible, then the function \[g\] is exclusive, which implies that there 's exactly one function \[g\] satisfying this property. That function \[g\] is then called the inverse of $f$, and is typically denoted as${f^{ - 1}}$.